1. Field of the Invention
The described invention relates to the field of communications. In particular, the invention relates to a method and apparatus for reducing echo in a full duplex transceiver system.
2. Description of Related Art
In a full duplex transceiver system, the output of the transmitter and the input of the receiver share the same path for connecting to an external data line. The transmitter""s transmission is thus received at the input of the receiver producing an xe2x80x9cechoxe2x80x9d, and a way of reducing this echo is employed so as not to interfere with the correct reception of signals by the receiver.
Various digital subscriber lines (DSL) as well as some Ethernet lines support full duplex transceiver systems. For example, the Gigabit Ethernet (IEEE specification 802.3) supports a full duplex transceiver system, and the 10 Gigabit Ethernet specification, although not yet adopted, is also likely to support a full duplex transceiver system.
One prior art method of reducing the echo is performed by subtracting a replica current from the receiver, as will be shown with respect to FIG. 1. The replica current is typically a predetermined fraction of the transmitter current.
As shown in FIG. 1, transmitter 19 provides currents +ITX and xe2x88x92ITX on input/output (I/O) lines 10 and 12, respectively. The I/O lines are also coupled to a receiver 20.
FIG. 1 shows only the input stage of the receiver 20, which comprises resistors 22, 24, 46, and 48, capacitors 60 and 62, and an operation amplifier (op amp) 30. Resistors 22 and 24 each have values M*RT and couple the I/O lines 10 and 12 to the op amp 30 via a differential circuit path that carries differential current IFB The inputs of the op amp 32 and 34 are coupled to differential replica current paths 42 and 44 that carry a differential current ITX/Nxe2x88x92(xe2x88x92ITX/N)=2 ITX/N.
Two feedback resistors RFB 46 and 48 couple the inputs 32 and 34 of the op amp 30 to its outputs 52 and 54, respectively. Capacitors 60 and 62 also couple the inputs 32 and 34 of the op amp 30 to its outputs 52 and 54.
The transmitter and receiver are generally on the same semiconductor chip and a termination resistor RT 70 is off-chip to match the impedance of a data line 80 such as a DSL or Gigabit Ethernet line. A transformer 90 couples the data line 80 to the I/O lines 10 and 12. A circuit board is typically used to mount the semiconductor chip, transformer 90, and resistor RT 70. A center tap 92 of the transformer 90 is coupled to the power supply of the circuit board. This center tap provides the differential current for the transmitter shown as +ITX and xe2x88x92ITX in FIG. 1.
For the circuit shown in FIG. 1, using the condition N=4M+1, it will now be shown that the dc output signal at the op amp output is just equal to the received signal from the external data line:
The transmitter voltage, in the absence of any received signal (i.e., for VRX=0), is given by the equations:
VTX=ITXRL, where the load across the transmitter is given by:                               R          L                =                              R            T                    ⁢                      "LeftDoubleBracketingBar"                          R              0                        "RightDoubleBracketingBar"                    ⁢          2          ⁢                      MR            T                                                            =                      (                                          2                ⁢                                  MR                  0                                ⁢                                  /                                ⁢                4                ⁢                M                            +              1                        )                          ,                              where            ⁢                          xe2x80x83                        ⁢                          R              T                                =                                    R              0                        .                              
The differential voltage VMX across the I/O lines 10 and 12 as shown in FIG. 1 is made up of the voltage component contributed by the transmitter and the voltage component contributed by the receiver:                               V          MX                =                              V            TX                    +                      V            RX                                                  =                                            (                                                2                  ⁢                                      MR                    0                                    ⁢                                      /                                    ⁢                  4                  ⁢                  M                                +                1                            )                        ⁢                          I              TX                                +                      V            RX                              
The differential current flowing through the resistors M RT 22 and 24 is given by the equation:                               I          FB                =                              V            MX                    ⁢                      /                    ⁢                      MR            T                                                            =                                                    (                                  2                  ⁢                                      /                                    ⁢                                      (                                                                  4                        ⁢                        M                                            +                      1                                        )                                                  )                            ⁢                              I                TX                                      +                          (                                                V                  RX                                ⁢                                  /                                ⁢                                  MR                  0                                            )                                      ,                              where            ⁢                          xe2x80x83                        ⁢                          R              T                                =                                    R              0                        .                              
The differential replica current leaving the receiver through replica current paths 42 and 44 is given by the equation:
IRD=xe2x88x922ITX/N
The output of the op amp 30 is:                               V          OUT                =                              [                                          I                FB                            +                              I                RD                                      ]                    ⁢                      R            FB                                                  =                              [                                                            (                                      2                    ⁢                                          /                                        ⁢                                          (                                                                        4                          ⁢                          M                                                +                        1                                            )                                                        )                                ⁢                                  I                  TX                                            -                                                (                                      2                    ⁢                                          /                                        ⁢                    N                                    )                                ⁢                                  I                  TX                                            +                              (                                                      V                    RX                                    ⁢                                      /                                    ⁢                                      MR                    0                                                  )                                      ]                    ⁢                      R            FB                                                            =                      V            RX                          ,                              where            ⁢                          xe2x80x83                        ⁢            N                    =                                                    4                ⁢                M                            +                              1                ⁢                                  xe2x80x83                                ⁢                and                ⁢                                  xe2x80x83                                ⁢                                  R                  FB                                                      =                          MR              0                                          
Thus, the output of the op amp 30 is just the received signal from the external data line. However, although the above description holds for a dc signal, this may not be the case for all frequencies.